Prob with understanding Bezier Curve Example

Discussion in 'Homework Help' started by zulfi100, Apr 13, 2014.

Jun 7, 2012
320
0
Hi,
I am trying to understand a problem related to Bezier Curve. I have downloaded it from a site.
The Question is attached in the file. The solution is also attached.
It has used the eq:
P(u) = P0(1-u)[SUP]3[/SUP] + 3P[SUB]1[/SUB]u(1-u)[SUP]2[/SUP]+3P[SUB]2[/SUB]u[SUP]2[/SUP](1-u)+P[SUB]3[/SUB]u[SUP]3[/SUP]

If the forum does not support the tags plz see the attached file.
For
P(1/4) =27/64P0+27/64P1+9/64P2+1/64P3

After that the solution shows the answer : [2.156 2.563 0].
I cant understand how to get this answer.Somebody plz guide me how can i get these values.

Zulfi.

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• Sol of Bezeir Curve Prob Part2.png
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Last edited: Apr 13, 2014
2. WBahn Moderator

Mar 31, 2012
17,715
4,788
What part don't you understand?

How to get the equation on the middle slide?

How those equations become the equations on the final slide?

How to evaluate the equations on the final slide?

As usual, we have to beg you to show some work so that we have a hope of figuring out where you are going off the rails.

Finally, you have been on this forum long enough that it is well past time that you learn how to use the LaTex equation rendering tool supported by this forum instead of always posting stuff that won't render because it is using some other rendering engine's tags.

After all, how hard is it to take your equation and modify it to:

P(u) = P_0(1-u)^3 + 3P_1u(1-u)^2+3P_2u^2(1-u)+P_3u^3

(note that I fixed your P0 error)

so that it will render between tex tags as

$
P(u) = P_0(1-u)^3 + 3P_1u(1-u)^2+3P_2u^2(1-u)+P_3u^3
$

Jun 7, 2012
320
0
Hi,
Thanks for your help. I cant understand how to substitute the values. The equation is in the form P0 , P1, P2, and P3 but values are in the form of x & y. Plz guide me.

Zulfi.

4. WBahn Moderator

Mar 31, 2012
17,715
4,788
We've been through this before. Each point has three components -- x, y, z --. The equations written in terms of P0 and such simply represents three separate equations, one in x, one in y and one in z. So write three equations. In the first, any place you see P2, replace it with x2. In the second, replace P2 with y2. In the third, replace P2 with z2. Do the same for the other points in the equation.

Jun 7, 2012
320
0
Hi,
Thanks for your guidance. I have solved it.
P(0) = P0
For x-component:
P(0) = x0 = 2
For y-component
P(0) = y0 = 2
For z-component
P(0) = z0 = 0
Because no values are provided for z-component
P(1/4) =27/64P0+27/64P1+9/64P2+1/64P3
For x-component
P(1/4) = 27/64x0 + 27/64x1 + 9/64x2 + 1/64x3
= 27/64(2) + 27/64(2) + 9/64(3) + 1/64(3)
=0.843 + 0.843 + .421 + .046
= 2.153875
For y-component
P(1/4) = 27/64y0 + 27/64y1 + 9/64y2 + 1/64y3
= 27/64(2) + 27/64(3) + 9/64(3) + 1/64(2)
= 0.843 + 1.265 + 0.421 + 0.031 = 2.56
For Z-component:
No values are provided for z-component, so
P(1/4) = 0
[2.1538, 2.56, 0]
P(1/2) = 1/8P0 + 3/8 P1 + 3/8P2 + 1/8P3
For x-component
P(1/2) = 1/8x0 + 3/8x1 + 3/8x2 + 1/8x3
= 1/8(2) + 3/8(2) + 3/8 (3) + 1/8(3)
= 0.25 + 0.75 + 1.125 + 0.375= 2.5
For y-component
P(1/2) = 1/8y0 + 3/8y1 + 3/8y2 + 1/8y3
=1/8(2) + 3/8(3) + 3/8(3) + 1/8(2)
= 0.25 + 1.125 + 1.125 + 0.25 = 2.75
For z-component
No values are provided for z-component, so its 0
[2.5, 2.75, 0]
P(3/4) = 1/64 x0 + 9/64 x1 + 27/64 x2 + 27/64 x3
= 1/64(2) + 9/64 (2) + 27/64 (3) + 27/64 (3)
= .0312 + 0.281 + 1.265 + 1.265 = 2.842
P(3/4) = 1/64 y0 + 9/64 y1 + 27/64 y2 + 27/64 y3
= 1/64 (2) + 9/64 (3) + 27/64 (3) + 27/64 (2) =
= 0.0312 + 0.421 + 1.265 + 0.843 = 2.56
[2.842, 2.56, 0]
P(1) = P3
For x-component
P(1) = x3= 3
For y-component
P(1) = y3=2
For z-component
No values are provided z-component, so its value is 0
[3, 2, 0]